What is the Formula for ln ex/ex?

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Homework Help Overview

The discussion revolves around the differentiation of the expression y = ln(e^x)/e^x, where participants are exploring the appropriate methods and rules for differentiation in the context of calculus.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the differentiation of the expression, with some suggesting the use of the quotient rule. Others question the simplification of ln(e^x) and its implications for the differentiation process.

Discussion Status

The conversation is active, with participants providing guidance on differentiation techniques and exploring the simplification of the expression. There is a focus on understanding the application of the quotient rule and the simplification of logarithmic expressions.

Contextual Notes

Some participants express confusion about the cancellation of terms in the differentiation process, indicating a need for clarification on the simplification steps involved.

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[SOLVED] need formula or how to...

y= ln ex/ex (x is the power)
 
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uhm use the "^" to represent powers, for example, x squared is written as x^2

and please state your question better
 
rock.freak667 said:
uhm use the "^" to represent powers, for example, x squared is written as x^2

and please state your question better

Differentiate:

y= ln e^x/e^x


Thanks!
 
So the question is to differentiate

[tex]y=\frac{ln e^x}{e^x}[/tex] ?


if it is...what rule do you think you would use when you need to differentiate something of the form [itex]y=\frac{u}{v}[/itex] ?
 
quotient rule.
which is: first, derivative of second, minus second, derivative of first, divided by first squared. (but, I'm still stuck. could you work it out for me?)
 
Last edited:
Well firstly...lne^x can be simplified to xlne which is just x

so you really want to find [tex]\frac{d}{dx}(\frac{x}{e^x})[/tex]

so if u= x => [itex]\frac{du}{dx}=1[/itex]
and v=e^x => [itex]\frac{dv}{dx}=e^x[/itex]

the formula is
[tex]\frac{d}{dx}(\frac{u}{v}) =\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}[/tex]so then you'll have
[tex]\frac{d}{dx}(\frac{x}{e^x}) = \frac{(e^x)(1) -(x)(e^x)}{(e^x)^2}[/tex]

then you will simplify it
 
thanks a lot. you made it very easy to understand. only part i didn't get is it says answer should be 1-x/e^x. what cancels out the power in the denominator?


Thanks again!
 
...well the e^x in the numerator can be factored out and it will cancel with an e^x in the denominator...so giving you your answer
 
oh, i see. thank you very much.. :)
 

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